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Chapter 17: Problem 60

Suppose a star experiences an outburst in which its surface temperature doubles but its average density (its mass divided by its volume) decreases by a factor of 8 . The mass of the star stays the same. By what factors do the star's radius and luminosity change?

Short Answer

Expert verified
The radius of the star changes by a factor of 2, while the luminosity changes by a factor of 64.

Step by step solution

01

Express the radius through density

From the volume and density formulas, one can isolate the radius as \( R = \left( \frac{3M}{4\pi\rho} \right) ^{1/3} \).
02

Calculate the change in radius

Since the density decreases by a factor of 8 and the mass is constant, the radius will change by the cube root of 8 which is 2. The radius increases by a factor of 2.
03

Use the luminosity formula

The luminosity of the star is given by \( L = 4\pi R^2 \sigma T^4 \).
04

Calculate the change in luminosity

The surface temperature doubles, thus the temperature term in the luminosity equation \( T^4 \) increases by a factor of \( 2^4 = 16 \). The radius increases by a factor of 2, therefore, the term \( R^2 \) in the luminosity equation increases by a factor of \( 2^2 = 4 \). So, the luminosity increases by a factor of \( 16 \times 4 = 64 \).

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